Analytic Functions with Isolated Zeros of Order k

[Euge]

Suppose $f$ is analytic in an open set $\Omega \subset \mathbb{C}$. Let $z_0\in \mathbb{C}$ and $r > 0$ such that the closed disk $\mathbb{D}_r(z_0) \subset \Omega$. If $f$ has a zero of order $k$ at $z = z_0$ and no other zeros inside $\mathbb{D}_r(z_0)$, show that there an open disk $D$ centered at the origin such that for all $\alpha\in D$, $f$ takes on the value $\alpha$ exactly $k$ times, counting multiplicity.

 

[Office_Shredder]

Is the final disk really centered at 0 and not $z_0$?

Edit; never mind, I'm bad at reading.

 

[julian]

Spoiler

We wish to show that

\begin{align*}
f (z) - \alpha = 0
\end{align*}

happens exactly $k$ times, counting multiplicity inside $\mathbb{D}_r (z_0)$ for all $\alpha$ such that $|\alpha| < R$ for some $R >0$.

Rouche's theorem:

"Let $f$ and $g$ be analytic in a simply connected domain $U \in \mathbb{C}$. Let $C$ be a simple closed contour in $U$. If $|f(z)| > |g(z)|$ for every $z$ on $C$, then the functions $f(z)$ and $f(z) + g(z)$ have the same number of zeros, counting multiplicities, inside $C$."

Take $C$ to be the circle centred at $z_0$ with radius $r$. Note $|f(z)| \not= 0$ on $C$. Let $R = \min_C |f(z)|$ and define an open disk $D$ about the origin of radius $R$. For $\alpha \in D$, write $g(z) = - \alpha$. Then $|f(z)| > |g(z)|$ for every $z$ on $C$. By Rouche's theorem $f(z)$ and $f(z) + g(z)$ have the same number of zeros, counting multiplicities, inside $C$. Therefore, $f$ takes on the value $\alpha$ exactly $k$ times, counting multiplicity, inside $\mathbb{D}_r (z_0)$.